**Elastic Beams in Three Dimensions**

**Elastic Beams in Three Dimensions**

**Lars Andersen and Søren R.K. Nielsen**

**Lars Andersen and Søren R.K. Nielsen**

**ISSN 1901-7286**

Department of Civil Engineering Structural Mechanics

**DCE Lecture Notes No. 23**

**Elastic Beams in Three Dimensions**

**Elastic Beams in Three Dimensions**

by

### Lars Andersen and Søren R.K. Nielsen

August 2008

c

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Published 2008 by Aalborg University

Department of Civil Engineering Sohngaardsholmsvej 57, DK-9000 Aalborg, Denmark

Printed in Denmark at Aalborg University ISSN 1901-7286 DCE Lecture Notes No. 23

**Preface**

This textbook has been written for the course Statics IV on spatial elastic beam structures given at the 5th semester of the undergraduate programme in Civil Engineering at Aalborg Uni-versity. The book provides a theoretical basis for the understanding of the structural behaviour of beams in three-dimensional structures. In the course, the text is supplemented with labora-tory work and hands-on exercises in commercial structural finite-element programs as well as MATLAB. The course presumes basic knowledge of ordinary differential equations and struc-tural mechanics. A prior knowledge about plane frame structures is an advantage though not mandatory. The authors would like to thank Mrs. Solveig Hesselvang for typing the manuscript.

**Contents**

**1** **Beams in three dimensions** **1**

1.1 Introduction . . . 1

1.2 Equations of equilibrium for spatial beams . . . 1

1.2.1 Section forces and stresses in a beam . . . 3

1.2.2 Kinematics and deformations of a beam . . . 5

1.2.3 Constitutive relations for an elastic beam . . . 10

1.3 Differential equations of equilibrium for beams . . . 12

1.3.1 Governing equations for a Timoshenko beam . . . 13

1.3.2 Governing equations for a Bernoulli-Euler beam . . . 14

1.4 Uncoupling of axial and bending deformations . . . 15

1.4.1 Determination of the bending centre . . . 15

1.4.2 Determination of the principal axes . . . 21

1.4.3 Equations of equilibrium in principal axes coordinates . . . 25

1.5 Normal stresses in beams . . . 28

1.6 The principle of virtual forces . . . 29

1.7 Elastic beam elements . . . 33

1.7.1 A plane Timoshenko beam element . . . 34

1.7.2 A three-dimensional Timoshenko beam element . . . 41

1.8 Summary . . . 44

**2** **Shear stresses in beams due to torsion and bending** **45**
2.1 Introduction . . . 45

2.2 Homogeneous torsion (St. Venant torsion) . . . 46

2.2.1 Basic assumptions . . . 47

2.2.2 Solution of the homogeneous torsion problem . . . 48

2.2.3 Homogeneous torsion of open thin-walled cross-sections . . . 57

2.2.4 Homogeneous torsion of closed thin-walled cross-sections . . . 59

2.3 Shear stresses from bending . . . 67

2.3.1 Shear stresses in open thin-walled cross-sections . . . 69

2.3.2 Determination of the shear centre . . . 75

2.3.3 Shear stresses in closed thin-walled sections . . . 82

2.4 Summary . . . 91

**C**

**HAPTER**

**1**

**Beams in three dimensions**

This chapter gives an introduction is given to elastic beams in three dimensions. Firstly, the
equations of equilibrium are presented and then the classical beam theories based on
Bernoulli-Euler and Timoshenko beam kinematics are derived. The focus of the chapter is the flexural
de-formations of three-dimensional beams and their coupling with axial dede-formations. Only a short
introduction is given to torsional deformations, or twist, of beams in three dimensions. A full
de-scription of torsion and shear stresses is given in the next chapters. At the end of this chapter, a
stiffness matrix is formulated for a three-dimensional Timosheko beam element. This element can
be used for finite-element analysis of elastic spatial frame structures.
**1.1**

**Introduction**

In what follows, the theory of three-dimensional beams is outlined.

**1.2**

**Equations of equilibrium for spatial beams**

An initially straight beam is considered. When the beam is free of external loads, the beam
occupies a so-called referential state. In the referential state the beam is cylindrical with the
lengthl*, i.e. the cross-sections are everywhere identical. The displacement and rotation of the*
beam is described in a referential(x, y, z)-coordinate system with base unit vectors{i,j,k}, the
originOplaced on the left end-section, and thex-axis parallel with the cylinder and orientated
into the beam, see Fig. 1–1. For the time being, the position ofOand the orientation of they
-andz-axes may be chosen freely.

The beam is loaded by a distributed load per unit length of the referential scale defined by the vector fieldq=q(x)and a distributed moment load vector per unit lengthm =m(x). A differential beam element of the lengthdxis then loaded by the external force vectorqdx and external moment vectormdxas shown in Fig. 1–1. The length of the differential beam element may change during deformations due to axial strains. However, this does not affect the indicated load vectors which have been defined per unit length of the referential state. Measured in the

(x, y, z)-coordinate system,qandmhave the components

q= qx qy qz , m= mx my mz . (1–1)

i j k x y z idx qdx mdx dx l −M M+dM −F F+dF

**Figure 1–1**Beam in referential state.

As a consequence of the external loads, the beam is deformed into the so-called current state where the external loads are balanced by an internal section force vectorF = F(x) and an internal section moment vectorM=M(x). These vectors act on the cross-section with the base unit vectoriof thex-axis as outward directed normal vector. With reference to Fig. 1–2, the components ofFandMin the(x, y, z)-coordinate system are:

F= N Qy Qz , M= Mx My Mz (1–2)

Here,N =N(x)*is the axial force, whereas the components*Qy=Qy(x)andQz=Qz(x)

*sig-nify the shear force components in the*y- andz-directions. The axial componentMx=Mx(x)of

*the section moment vector is denoted the torsional moment. The components*My=My(x)and
Mz=Mz(x)in they- andz*-directions represent the bending moments. The torsional moment*

is not included in two-dimensional beam theory. However, in the design of three-dimensional frame structures, a good understanding of the torsional behaviour of beams is crucial.

Assuming that the displacements remain small, the equation of static equilibrium can be established in the referential state. With reference to Fig. 1–1, the left end-section of the element is loaded with the section force vector−Fand the section moment vector−M. At the right end-section, these vectors are changed differentially intoF+dFandM+dM, respectively. Force equilibrium and moment equilibrium formulated at the point of attack of the section force vector−Fat the left end-section then provides the following equations of force and moment

x y z Mx My Mz N Qy Qz

**Figure 1–2**Components of the section force vector and the section moment vector.

equilibrium of the differential beam element: −F+F+dF+qdx=0⇒ dF dx +q=0 (1–3a) −M+M+dM+idx×(F+dF) +mdx=0⇒ dM dx +i×F+m=0 (1–3b)

From Eqs. (1–1) and (1–2) follows that Eqs. (1–3a) and (1–3b) are equivalent to the following component relations: dN dx +qx= 0, dQy dx +qy= 0, dQz dx +qz= 0, (1–4a) dMx dx +mx= 0, dMy dx −Qz+my= 0, dMz dx +Qy+mz= 0. (1–4b) At the derivation of Eq. (1–4b), it has been utilised that

i×F=i×(Ni+Qyj+Qzk) =Ni×i+Qyi×jQzi×k= 0i−Qzj+Qyk. (1–5)

Hence,i×Fhas the components{0,−Qz, Qy}. It is noted that a non-zero normal-force

compo-nent is achieved when the moment equilibrium equations are formulated in the deformed state. This may lead to coupled lateral-flexural instability as discussed in a later chapter.

**1.2.1**

**Section forces and stresses in a beam**

On the cross-section with the outward directed unit vector co-directional to thex-axis, the normal stressσxxand the shear stressesσxyandσxzact as shown in Fig. 1–3. These stresses must be

statically equivalent to the components of the force vectorFand the section moment vectorM as indicated by the following relations:

N = Z A σxxdA, Qy = Z A σxydA, Qz= Z A σxzdA, (1–6a) Mx= Z A (σxzy−σxyz)dA, My= Z A zσxxdA, Mz=− Z A yσxxdA. (1–6b) x y z dA Mx My Mz N Qy Qz σxx σxy σxz

**Figure 1–3**Stresses and stress resultant on a cross-section of the beam.

x y z dx dy dz σxx σyy σzz σxy σyx σxz σzx σyz σzy

**Figure 1–4**Components of the stress tensor.

On sections orthogonal to they- andz-axes, the stresses{σyy, σyx, σyz}and{σzz, σzx, σzy}

act as shown in Fig. 1–4. The first index indicates the coordinate axis co-directional to the outward normal vector of the section, whereas the second index specifies the direction of action of the stress component. The stresses shown in Fig.1–4 form the components of the stress tensor

σin the(x, y, z)-coordinate system given as
σ_{=}
σxx σyx σzx
σxy σyy σzy
σxz σyz σzz
. (1–7)

Moment equilibrium of the cube shown in Fig. 1–4 requires that

σxy =σyx, σxz=σzx, σyz=σzy. (1–8)

Hence,σis a symmetric tensor.

**1.2.2**

**Kinematics and deformations of a beam**

The basic assumption in the classical beam theory is that a cross-section orthogonal to thex-axis at the coordinatexremains plane and keeps its shape during deformation. In other words, the cross-section translates and rotates as a rigid body. Especially, this means that Poisson contrac-tions in the transverse direction due to axial strains are ignored. Hence, the deformed position of the cross-section is uniquely described by a position vectorw =w(x)and a rotation vector

θ_{=}θ_{(}_{x}_{)}_{with the following components in the}_{(}_{x, y, z}_{)}_{-coordinate system:}

w= wx wy wz , θ= θx θy θz . (1–9)

Further, only linear beam theory will be considered. This means that the displacement compo-nentswx, wy andwzin Eq. (1–9) all small compared to the beam lengthl. Further the rotation

componentsθx, θyandθzare all small. Especially, this means that

sinθ≃tanθ≃θ, (1–10)

whereθrepresents any of the indicated rotation components measured in radians. The various displacement and rotation components have been illustrated in Fig. 1–5. The rotation component around thex-axis is known as the twist of the beam.

Now, a material point on the cross-section with the coordinates(x, y, z)in the referential state achieves a displacement vectoru = u(x, y, z)with the components{ux, uy, uz} in the (x, y, z)-coordinate system given as (see Fig. 1–5):

ux(x, y, z) =wx(x) +zθy(x)−yθz(x), (1–11a)

uy(x, y, z) =wy(x)−zθx(x), (1–11b)

uz(x, y, z) =wz(x) +yθx(x). (1–11c)

It follows that the displacement of any material point is determined if only the 6 components of
w(x)andθ_{(}_{x}_{)}_{are known at the beam coordinate}_{x}_{. Hence, the indicated kinematic constraint}

reduces the determination of the continuous displacement fieldu=u(x, y, z)to the determina-tion of the 6 deformadetermina-tion componentswx =wx(x),wy =wy(x),wz = wz(x),θx = θx(x), θy=θy(x)andθz=θz(x)of a single spatial coordinate along the beam axis.

x x x y y y z z z dwy dx dwz dx wx wx wy wz θx −θy θz

**Figure 1–5**Deformation components in beam theory.

The strains conjugated to σxx,σxyandσxz are the axial strainεxxand the angular strains γxy= 2εxyandγxz= 2εxz. They are related to displacement components as follows:

εxx= ∂ux ∂x = dwx dx +z dθy dx −y dθz dx, (1–12a) γxy= ∂ux ∂y + ∂uy ∂x = dwy dx −z dθx dx −θz(x), (1–12b) γxz= ∂ux ∂z + ∂uz ∂x = dwy dx +y dθx dx +θy(x). (1–12c)

From Eq. (1–12) follows thatγxy =γxy(x, z)is independent ofyas a consequence of the

presumed plane deformation of the cross-section. Then, the shear stressσxy =σxy(x, z)must

also be constant over the cross-section. Especially,σxy 6= 0at the upper and lower edge of the

cross-section as illustrated in Fig. 1–6a. However, if the cylindrical surface is free of surface shear tractions, thenσyx = 0 at the edge. Hence,σxy 6= σyx in contradiction to Eq. (1–8).

In reality σxy = 0 at the edges, corresponding to γxy = 0. This means that the deformed

cross-section forms a right angle to the cylindrical surface as shown in Fig. 1–6b.

The displacement fields Eq. (1–11) are only correct for beams with cross-sections which are circular symmetric around thex-axis. In all other cross-sections, the torsional momentMxwill

x x y y z z dwy dx dwy dx wx wx wy wy σxy σxy σyx γxy (a) (b)

**Figure 1–6**Shear stresses on deformed beam section: (a) Deformation of cross-section in beam theory and (b) real
deformation of cross-section.

induce an additional non-planar displacement in thex-axis, which generally can be written in the formux(x, y, z) = ω(y, z)dθx/dx. This is illustrated in Fig. 1–7. Hence, the final expression

for the axial displacement reads

ux(x, y, z) =wx(x) +zθy(x)−yθz(x) +ω(y, z)dθx

dx. (1–13)

The expressions foruyanduzin Eq. (1–11) remain unchanged, andω(y, z)*is called the *

*warp-ing function. Whereas* y and z in Eq. (1–13) may be considered as shape functions for the
deformations caused by the rotationsθz(x)andθy(x), the warping function is a shape function
defining the axial deformation of the cross-section from the rotation component. The definition
and determination of the warping function is considered in a subsequent section.

Deformed bottom flange

Deformed top flange Undeformed state

Section A–A Section B–B A A B B

**Figure 1–7**Warping deformations in an I-beam induced by homogeneous torsion. The cross sections A–A and B–B are
shown with the top flange on the left and the bottom flange on the right.

As a consequence of the inclusion of the warping, the strain components in Eq. (1–12) are
modified as follows:
εxx=
∂ux
∂x =
dwx
dx +z
dθy
dx −y
dθz
dx +ω
d2_{θ}
x
dx2 , (1–14a)
γxy=∂ux
∂y +
∂uy
∂x =
dwy
dx −θz+
∂ω
∂y −z
dθx
dx, (1–14b)
γxz=
∂ux
∂z +
∂uz
∂x =
dxz
dx +θy+
_{∂ω}
∂z +y
_{dθ}
x
dx. (1–14c)

x x y y z z dwy dx dwz dx wx wx wy wz −θy θz

**Figure 1–8**Kinematics of Bernoulli-Euler beam theory.

Bernoulli-Euler beam kinematics presumes that the rotated cross-section is always orthogo-nal to the deformed beam axis. This involves the following additioorthogo-nal kinematical constraints on the deformation of the cross-section (see Fig. 1–8):

θy=− dwz

dx , θz= dwy

dx . (1–15)

Assuming temporarily thatθx ≡ 0 *in bending deformations, i.e. disregarding the twist of the*

beam, Eqs. (1–14) and (1–15) then provide:

γxy=γxz= 0. (1–16)

Equation (1–16) implies that the shear stresses areσxy =σxz = 0, and in turn that the shear

forces becomeQy =Qz= 0, cf. Eq. (1–6). However, non-zero shear forces are indeed present

in bending of Bernoulli-Euler beams. The apparent paradox is dissolved by noting that the shear forces in Bernoulli-Euler beam theory cannot be derived from the kinematic condition, but has to be determined from the static equations.

The development of the classical beam theory is associated with names like Galilei (1564–1642), Mariotte (1620–1684), Leibner (1646–1716), Jacob Bernoulli (1654–1705), Euler (1707–1783), Coulomb (1736–1806) and Navier (1785–1836), leading to the mentioned Bernoulli-Euler beam based on the indicated kinematic constraint. The inclusion of transverse shear deformation was proposed in 1859 by Bresse (1822–1883) and extended to dynamics in 1921 by Timoshenko (1878– 1972). Due to this contribution, the resulting beam theory based on the strain relations Eq. (1–12), is referred to as Timoshenko beam theory (Timoshenko 1921).

The first correct analysis of torsion in beams was given by St. Venant (1855). The underlying assumption was that dθx/dxin Eq. (1–13) was constant, so the warping in all cross-sections

become identical. Then, the axial strainεxxfrom torsion vanishes and the distribution of the

shear strainsγxyandγxzare identical in all sections. Because of this, St. Venant torsion is also

Whenever the twist or the warping is prevented at one or more cross-sections,dθx/dxis no

longer constant as a function of x. Hence, axial strains occur and, as a consequence of this,
axial stresses arise and the shear strains and shear stresses are varying along the beam. These
phenomena were systematically analysed by Vlasov (1961) for thin-walled beams, for which
*reason the resulting theory is referred to as Vlasov torsion or non-homogeneous torsion. Notice*
that the shear stresses from Vlasov torsion have not been included in the present formulation.
These will be considered in a subsequent chapter.

Seen from an engineering point of view, the primary advantage of Vlasov torsion theory is that it
*explains a basic feature of beams, namely that prevention of warping leads to a much stiffer *

*struc-tural elements than achieved in the case of homogeneous warping, i.e. a given torsional moment*

will induce a smaller twist. Warping of the cross-section may, for example, be counteracted by the inclusion of a thick plate orthogonal to the beam axis and welded to the flanges and the web. The prevention of torsion in this manner is particularly useful in the case of slender beams with open thin-walled cross-sections that are prone to coupled flexural–torsional buckling. Obviously, Vlasov torsion theory must be applied for the analysis of such problems as discussed later in the book.

Next, the deformation of the cross-section may be decomposed into bending and shear com-ponents. The bending components are caused by the bending momentsMyandMzand deform

as a Bernoulli-Euler beam. Hence, the bending components are causing the rotationsθyandθz

of the cross-section. The shear components are caused by the shear forcesQy andQz. These

cause the angular shear strainsγxyandγxzwithout rotating the cross-section. Further, the

dis-placement of the beam axis in shear takes place without curvature. Hence, the curvature of the beam axis is strictly related to the bending components, see Fig. 1–9.

With reference to Fig. 1–10, the radii of curvatures ry andrz are related to the rotation

incrementsdθzand−dθyof the end-sections in the bending deformations of a differential beam

element of the lengthdxas follows rydθz=dx −rzdθy =dx ⇒ κy=−1/rz=dθy/dx κz= 1/ry=dθz/dx (1–17)

Here,κyandκzdenote the components of the curvature vectorκof thex-axis. Especially, for a

Bernoulli-Euler beam the curvature components become, cf. Eq. (1–15),
κy=−d
2_{w}
z
dx2 , κz=
d2_{w}
y
dx2 . (1–18)

From Eqs. (1–14) and (1–17) follows that the axial strain may be written as εxx(x, y, z) =ε(x) +zκy(x)−yκz(x) +ω(y, z)d

2_{θ}

x

dx2, (1–19)

whereε(x)denotes the axial strain of the beam along thex-axis given as ε(x) = dwx

dx . (1–20)

Here, ε(x), κy(x)andκy(x) *define the axial strain and curvatures of the beam axis, i.e. the*

### =

### +

x x x y y y z z z Mz Qy γxy γxy**Figure 1–9**Decomposition of cross-section deformation into bending and shear components.

x x y y z z ds≈dx ds≈dx ry rz wy wz −dθy dθz

**Figure 1–10**Definition of curvature.

**1.2.3**

**Constitutive relations for an elastic beam**

In what follows we shall refer toN(x),Qy(x),Qz(x),Mx(x),My(x)andMz(x)*as generalised*

*stresses. These are stored in the column matrix*

σ_{(}_{x}_{) =}
N(x)
Qy(x)
Qz(x)
Mx(x)
My(x)
Mz(x)
. (1–21)

The internal virtual work of these quantities per unit length of the beam is given as
δω=N δε+Qyδγxy+Qzδγxz+Mxδθx+Myδκy+Mzδκz=σTδε, (1–22)
where
ε_{(}_{x}_{) =}
ε(x)
γxy(x)
γxz(x)
θx(x)
κy(x)
κz(x)
. (1–23)

The components ofε_{(}_{x}_{)}* _{are referred to as the generalised strains. The components of}*σ

_{(}

_{x}

_{)}

_{and}ε

_{(}

_{x}

_{)}

_{are said to be virtual work conjugated because these quantities define the internal virtual}

work per unit length of the beam.

LetEandGdenote the elasticity modulus and the shear modulus. Then, the normal stress σxxand the shear stressesσxyandσxzmay be calculated from Eq. (1–14) as follows:

σxx=Eεxx=E
dwx
dx +z
dθy
dx −y
dθz
dx +ω(y, z)
d2_{θ}
x
dx2
, (1–24a)
σxy =Gγxy=G
dwy
dx −θz+
∂ω
∂y −z
dθx
dx
, (1–24b)
σxz=Gγxz=G
dw2
dx +θy+
∂ω
∂z +y
dθx
dx
. (1–24c)

By integration over the cross-sectional area, it then follows that
N =E
Adwx
dx +Sy
dθy
dx −Sz
dθ2
dx +Sω
d2_{θ}
x
dx2
, (1–25a)
Qy=G
Ay
_{dw}
y
dx −θz
+Ry
dθx
dx
, (1–25b)
Qz=G
Az
_{dw}
z
dx +θy
+Rz
dθx
dx
, (1–25c)
Mx=G
Sz
dwz
dx +θy
−Sy
dwy
dx −θz
+Kdθx
dx
, (1–25d)
My=E
Sy
dwx
dx +Iyy
dθy
dx −Iyz
dθz
dx +Iωz
dθx
dx
, (1–25e)
Mz=E
−Sz
dwx
dx −Iyz
dθy
dx +Izz
dθz
dx −Iωy
dθx
dx
, (1–25f)

whereAy,Az,Ry,Rz,Sy,Sz,Sω,K,Iyy,Izz,Iyz =Izy,Iωy andIωzare cross-sectional (or

geometrical) constants identified as: A=

Z A

Ry = Z A ∂ω ∂y −z dA, Rz= Z A ∂ω ∂z +y dA, (1–26b) Sy= Z A zdA, Sz= Z A ydA, Sω= Z A ωdA, (1–26c) Iyy= Z A z2dA, Izz= Z A y2dA, (1–26d) Iyz = Z A yzdA, Iωy= Z A ωydA, Iωz= Z A ωzdA, (1–26e) K= Z A y2+z2+y∂ω ∂z −z ∂ω ∂y dA. (1–26f)

Here,Ais the cross-sectional area, whereasAy andAzsignify the so-called shear areas. Beam

theory presumes a constant variation of the shear stresses in bending, whereas the actual variation is at least quadratic. The constant variation results in an overestimation of the stiffness against shear deformations, which is compensated by the indicated shear reduction factorsαyandαz. If

the actual distribution of the shear stresses is parabolic, these factors becomeαy =αz = 5/6.

For anI-profile, the shear area is approximately equal to the web area.

ForGAy → ∞we haveγxy = Qy/(GAy) = 0. Bernoulli-Euler beam theory is

charac-terised byγxy = 0. Hence, Timoshenko theory must converge towards Bernoulli-Euler theory

for the shear areas passing towards infinity. The magnitude of the shear deformations in
propor-tion to the bending deformapropor-tions depends on the quantity(h/l)2_{, where}_{h}_{is the height and}_{l}_{is}
the length of the beam. This relation is illustrated in Example 1-3.

RyandRzare section constants which depend on the warping mode shapeω(y, z)as well

as the bending modes viayandz. Further, the section constantsSyandSzare denoted the static

moments around they- andz-axes. Sωspecifies a corresponding static moment of the warping

shape function.

IyyandIzz *signify the bending moments of inertia around the*y- andz-axes, respectively.
Iyz *is denoted the centrifugal moment of inertia, whereas* Iωy andIωz are the corresponding

*centrifugal moments of the warping shape function and the bending mode shapes.*

K*is the so-called torsion constant. This defines merely the torsional stiffness in St. Venant*
torsion. As mentioned above, the additional contribution to Mx from Vlasov torsion will be

considered in a subsequent section.

**1.3**

**Differential equations of equilibrium for beams**

In what follows, the governing differential equations for Timoshenko and Bernoulli-Euler beams are derived. At this stage, the twistθxand the torsional momentMxare ignored. With no further

assumptions and simplifications, Eq. (1–25) reduces to N My Mz Qy Qz = EA −ESy −ESz 0 0 ESy EIyy −EIyz 0 0 −ESz −EIyz EIzz 0 0 0 0 0 GAy 0 0 0 0 0 GAz dwx/dx dθy/dx dθz/dx dwy/dx−θz dwz/dx+θy . (1–27)

The coefficient matrix of Eq. (1–27) is symmetric. When formulated in a similar matrix format, the corresponding matrix in Eq. (1–25) is not symmetric. This is a consequence of the ignorance of the Vlasov torsion inMx.

**1.3.1**

**Governing equations for a Timoshenko beam**

Next, Eq. (1–27) is inserted into the equilibrium equations (1–4a) and (1–4b), which results in the following system of coupled ordinary differential equations for the determination ofwx,wy, wz,θyandθz: dN/dx dMy/dx dMz/dx dQy/dx dQz/dx = 0 Qz −Qy 0 0 − qx my mz qy qz ⇒ d dx EA ESy −ESz 0 0 ESy EIyy −EIyz 0 0 −ESz −EIyz EIzz 0 0 0 0 0 GAy 0 0 0 0 0 GAz dwx/dx dθy/dx dθz/dx dwy/dx−θz dwz/dx+θy = 0 0 0 0 0 0 0 0 0 GAz 0 0 0 −GAy 0 0 0 0 0 0 0 0 0 0 0 dwx/dx dθy/dx dθz/dx dθz/dx−θz dwz/dx+θy − qx my mz qy qz . (1–28)

Equation (1–28) specifies the differential equations for Timoshenko beam theory. These should
be solved with proper boundary condition at the end-sections of the beam. Letx0 denote the
*abscissa of any of the two end-sections, i.e.*x0= 0orx0=l, wherelis the length of the beam.
Atx=x0either kinematical or mechanical boundary conditions may be prescribed.

Kinematical boundary conditions mean that values ofwx,wy,wz,θyandθzare prescribed, wx(x0) = wx,0 wy(x0) = wy,0 wz(x0) = wz,0 θy(x0) = θy,0 θz(x0) = θz,0 , x0= 0, l, (1–29)

whereas mechanical boundary conditions imply the prescription ofN,Qy,Qz,MyandMz, N(x0) = N0 Qy(x0) = Qy,0 Qz(x0) = Qz,0 My(x0) = My,0 Mz(x0) = Mz,0 , x0= 0, l. (1–30)

In Eq. (1–30), the left-hand sides are expressed in kinematical quantities by means of Eq. (1–27). Of the 10 possible boundary conditions atx=x0specified by Eqs. (1–35) and (1–30), only 5 can be specified. The 5 boundary conditions atx0= 0andx0=lcan be selected independently from Eq. (1–35) and Eq. (1–30).

With given boundary conditions Eq. (1–28) can be solved uniquely for the 5 kinematic quan-titieswx,wy,wz,θy,θz, which make up the degrees of freedom of the cross-section. Although

an analytical solution may be cumbersome, a numerical integration is always within reach.

**1.3.2**

**Governing equations for a Bernoulli-Euler beam**

Next, similar differential equations are specified for a Bernoulli-Euler beam. At first the shear forcesQyandQzin the equations of equilibrium forMy andMzin Eq. (1–4b) are eliminated

by means of the 2nd and 3rd equations in Eq. (1–4a):

d2_{M}
y/dx2−dQz/dx+dmy/dx= 0
d2Mz/dx2+dQy/dx+dmz/dx= 0
⇒
d2_{M}
y/dx2+qz+dmy/dx= 0
d2Mz/dx2−qy+dmz/dx= 0.
(1–31)

Using the Bernoulli-Euler kinematical constraint Eq. (1–15), the constitutive equations for the resulting section forces may be written as

N
My
Mz
=
EA ESy −ESz
ESy EIyy −EIyz
−ESzz −EIyz EIzz
dwx/dx
−d2_{w}
z/dx2
d2_{w}
y/dx2
. (1–32)

Then, the equations of equilibrium Eq. (1–4a) and Eq. (1–31) may be recasted as the following system of coupled ordinary differential equations

d
dx
EAdwx
dx −ESy
d2_{w}
z
dx2 −ESz
d2_{w}
y
dx2
+qx= 0, (1–33a)
d2
dx2
ESy
dwx
dx −EIyy
d2_{w}
z
dx2 −EIyz
d2_{w}
y
dx2
+qz+
dmy
dx = 0, (1–33b)
d2
dx2
−ESz
dwx
dx +EIyz
d2_{w}
z
dx2 +EIzz
d2_{w}
y
dx2
+qy+
dmz
dx = 0. (1–33c)

The governing equations (1–33) should be solved with 5 of the same boundary conditions as indicated by Eqs. (1–35) and (1–30). The difference is thatθy(x0),θz(x0),Qy(x0)andQz(x0) are represented as, cf. Eqs. (1–4b) and (1–15),

−dwz(x0) dx =θy,0, dwz(x0) dx =θz,0, (1–34a) −dMz(x0) dx −mz(x0) =Qy,0, dMy(x0) dx +my(x0) =Qz,0. (1–34b)

With this in mind, the kinematic boundary conditions for Bernoulli-Euler beams are given in the form wx(x0) = wx,0 wy(x0) = wy,0 wz(x0) = wz,0 dwz(x0)/dx = θy,0 dwy(x0)/dx = θz,0 , x0= 0, l, (1–35)

whereas the mechanical boundary conditions defined in Eq. (1–30) are still valid.

**1.4**

**Uncoupling of axial and bending deformations**

Up to now the position of the originO and the orientation of the y- andz-axes in the cross-section have been chosen arbitrarily. As a consequence of this, the deformations from the axial force and the deformation from the bending momentsMy andMz will generally be coupled.

This means that the axial forceN referred to the originO will not merely induce a uniform displacementwxof the cross-section, but also non-zero displacementswy andwzofO as well

as rotationsθyandθz. Similarly, the bending momentMywill not merely cause a displacement wyand a rotationθyof the cross-section, but also a non-zero displacementwyand a rotationθz

in the orthogonal direction in addition to an axial displacementwy of the origin. The indicated

mechanical couplings are the reason for the couplings in the differential equations (1–28) and (1–33). The couplings may have a significant impact on the structural behaviour and stability of an engineering structure and the position of the origin for a given beam element as well as the orientation of the coordinate axes must be implemented correctly in a computational model.

In this section, two coordinate transformations will be indicated, in which the axial force
re-ferred to the new originB*, called the bending centre, only induces a uniform axial displacement*
over the cross-section. Similarly, the bending momentsMy andMz around the new rotatedy

-andz*-axes, referred to as the principal axes, will only induce the non-zero deformation *
compo-nents(wz, θy)and(wy, θz), respectively. Especially, the moments will induce the displacement
wx= 0of the bending centre,B.

**1.4.1**

**Determination of the bending centre**

The position of the bending centreB is given by the position vectorrB with the components {0, yB, zB}in the(x, y, z)-coordinate system. In order to determine the componentsyBandzB,

a translation of the(x, y, z)-coordinate system to a new(x′_{, y}′_{, z}′_{)}_{-coordinate system with origin}
in the yet unknown bending centre is performed (see Fig. 1–11). The relations between the new
and the old coordinates read

x=x′_{,} _{y}_{=}_{y}′_{+}_{y}

B, z=z′+zB, (1–36)

In the new coordinate system, the displacement ofB(the new origin) in thex′_{-direction (the}
new beam axis) becomes (see Fig. 1–11):

w′

θy, My θz, Mz θ′ y, My′ θ′ z, Mz′ N N′ y z y′ z′ yB zB O B rB

**Figure 1–11**Translation of coordinate system.

The axial strain of fibres placed on the new beam axis becomes
ε′_{(}_{x}′_{) =}dw′

dx′ =

dw′

dx =ε+zBκy−yBκz, (1–38)

where Eq. (1–17), Eq. (1–20) and Eq. (1–37) have been used. The components of the rotation
vectorθ_{of the cross-section are identical, i.e.}

θ′

x=θx, θ′y=θy, θz′ =θz. (1–39)

In turn, this means that the components of the curvature vectorκin the two coordinate systems are identical as well

κ′ y= dθ′ y dx′ = dθy dx =κy, κ ′ z= dθ′ z dx′ = dθz dx =κz. (1–40)

Further, the components of the section force vectorFin the two coordinate systems become
*identical, i.e.*

N′_{=}_{N,} _{Q}′

y=Qy Q′z=Qz. (1–41)

As a consequence of referring the axial forceN =N′_{to the new origin}_{B}_{, the components}
of the section vector in the(x′_{, y}′_{, z}′_{)}_{-coordinate system are related to the components in the}

(x, y, z)-coordinate as follows: M′

x=Mx, My′ =My−zBN, Mz′ =Mz+yBN. (1–42)

Equations (1–38) and (1–40) provide the following relation forεin terms ofε′_{,}_{κ}′

yandκz:
ε=ε′_{−}_{z}

Then the relation between{N, My, Mz} and{N′, My′, Mz′} and{ε, κy, κz} and{ε′, κ′y, κz}

may be specified in the following matrix formulation:

σ′_{=}_{A}T_{σ}_{,} _{(1–44a)}
ε_{=}Aε′_{,} _{(1–44b)}
where
σ _{=}
N
My
M_{z}
, ε =
ε
κy
κ_{z}
, (1–45a)
σ′_{=}
N′
M′
y
M′
z
, ε′ =
ε′
κ′
y
κ′
y
, (1–45b)
A =
1 −zB yB
0 1 0
0 0 1
. (1–45c)

The components{N, My, Mz}and{ε, κy, κz}ofσandεmay be interpreted as work conjugated

generalised stresses and strains.

With reference to Eq. (1–32), the constitutive relation betweenσandεis given as

σ_{=}Cε_{,} (1–46)

whereCdenotes the constitutive matrix,

C=E A Sy −Sz Sy Iyy −Iyz −Sz −Iyz Izz . (1–47)

Likewise, the constitutive relation in the(x′_{, y}′_{, z}′_{)}_{-coordinate system reads}

σ′_{=}_{C}′_{ε}′ _{(1–48)}

where the constitutive matrix has the form

C′ _{=}_{E}
A Sy′ −Sz′
Sy′ Iy′y′ −Iy′z′
−Sz′ −Iy′z′ Iz′z′
. (1–49)

Obviously, as given by Eqs. (1–47) and (1–49), the cross-sectional areaAis invariant to a rotation of the cross-section about thex-axis and a translation in they- andz-directions.

From Eqs. (1–44a), (1–44b) and (1–46) follows that

σ′_{=}_{A}T_{C}_{ε}_{=}_{A}T_{CA}_{ε}′ _{⇒}
C′ _{=}_{A}T_{CA}_{=}_{E}
1 0 0
−zB 1 0
yB 0 1
A Sy −Sz
Sy Iyy −Iyz
−Sz −Iyz Izz
1 −zB yB
0 1 0
0 0 1
⇒

C′ _{=}_{E}

A Sy−zBA −(Sz−yBA)

Sy−zBA Iyy−2zBSy+z2BA −Iyz+yBSy+zB(Sz−yBA) −(Sz−yBA) −Iyz+yBSy+zB(Sz−yBA) Izz−2yBSz+y2BA

.

(1–50)
The idea is now to use the translational coordinate transformation to uncouple the axial
defor-mations from the bending defordefor-mations. This requires thatSy′ =S_{z}′ = 0. Upon comparison

of Eq. (1–49) and Eq. (1–50), this provides the following relations for the deformation of the coordinates of the bending centre:

yB =Sz

A, zB=

Sy

A. (1–51)

WithyBandzBgiven by Eq. (1–51), the bending moments of inertia,Iy′y′andIz′z′, and the

centrifugal moment of inertia,Iy′z′, in the new coordinate system can be expressed in terms of

the corresponding quantities in the old coordinate system as follows:

Iy′y′=Iyy−2zB(AzB) +z2BA=Iyy−zB2A, (1–52a) Iz′z′ =Izz−2yB(AyB) +y2BA=Izz−yB2A, (1–52b) Iy′z′ =Iyz−yB(AzB)−zB(AyB) +yBzBA=Iyz−yBzBA. (1–52c)

*The final results in Eq. (1–52) are known as König’s theorem.*

x

y

z yB

O

**Figure 1–12**Single-symmetric cross-section.

If the cross-section is symmetric around a single line, and the y-axis is placed so that it
coincides with this line of symmetry, then the static momentSy*vanishes, i.e.*

Sy= Z

A

zdA= 0 (1–53)

As a result of this, the bending centreB will always be located on the line of symmetry in a single-symmetric cross-section, see Fig. 1–12. Obviously, if the cross-section is double symmet-ric, then the position ofBis found at the intersection of the two lines of symmetry.

**Example 1.1 Determination of bending and centrifugal moments of inertia of non-symmetric**

thin-walled cross-section

The position of the bending centre of the cross-section shown in Fig. A is determined along with the bending moments of inertiaIy′y′andIz′z′and the centrifugal moment of inertiaIy′z′.

x y z t t 2t a 2a 2a O

**Figure A** Thin-walled cross-section.

The(x, y, z)-coordinate system is placed as shown in Fig. A. Then, the following cross-sectional con-stants are calculated:

A= 2a·2t+ 2a·t+a·t= 7at, (a)
Sy= 2a·2t·a+ 2a·t· t
2+a·t·
a
2=
1
2(2t+ 9a) +ta, (b)
Sz= 2a·2t·t+ 2a·t·(2t+a) +a·t·
2t+ 2a+t
2
=1
2(21t+ 8a)ta, (c)
Iyy=
1
3·2t·(2a)
3_{+}1
3·2a·t
3_{+}1
3·t·a
3_{=} 1
3 2t
2_{+ 17}_{a}2
ta, (d)
Izz=1
3·2a·2a·(2t)
3_{+} 1
12(2a)
3
·t+ 2a·t·(2t+a)2+ 1
12·a·t
3_{+}_{a}
·t·
2t+ 2a+ t
2
2
=1
3 59t
2_{+ 54}_{ta}_{+ 20}_{a}2
ta, (e)
Iyz= 2a·2t·t·a+ 2a·t·(2t+a)·
1
2+a·t·
2t+ 2a+t
2
·a_{2} = 1
4 8t
2_{+ 25}_{ta}_{+ 4}_{a}2
ta. (f)
Here, use has been made of König’s theorem at the calculation of contributions toIyy,IzzandIyzfrom
the three rectangles forming the cross-section.

The coordinates of the bending centre follow from Eq. (1–51) and Eqs. (a) to (c). Thus,
yB=Sz
A =
3
2t+
4
7a, zB=
Sy
A =
1
7t+
9
14a. (g)
*(continued)*

x′ y′ z′ B 3 2t+ 4 7a 1 7t+149a

**Figure B** Position of bending centre in the thin-walled cross-section.

Subsequently, the moments of inertia around the axes of the(x′_{, y}′_{, z}′_{)}_{-coordinate system follow from}

Eq. (1–52), Eq. (1–53) and Eqs. (d) to (f):
Iy′y′ =
1
3(2t
2_{+ 17}_{a}2_{)}_{ta}
−
1
7t+
9
14a
2
·7ta= 1
84(44t
2
−108ta+ 233a2)ta, (h)
Iz′z′ =
1
3(59t
2_{+ 54}_{ta}_{+ 20}_{a}2_{)}_{ta}
−
3
2t+
4
7a
2
·7ta= 1
84(329t
2_{+ 504}_{ta}_{+ 368}_{a}2_{)}_{ta,}
(i)
Iy′z′=
1
4(8t
2_{+ 25}_{ta}_{+ 4}_{a}2_{)}_{ta}
−
1
7t+
9
14a)
3
2t+
4
7a
·7ta= 1
14(7t
2
−15ta−22a2)ta. (j)
Now, for a thin-walled cross-section the thickness of the flanges and the web is much smaller than the
widths of the flanges and the height of the web. In the present case this means thatt≪a. With this in
mind, Eqs. (g) to (j) reduce to

yB≃4
7a, zB≃
9
14a (k)
and
Iy′y′ ≃
233
84ta
3_{,} _{I}
z′z′≃
92
21ta
3_{,} _{I}
y′z′≃ −
11
7ta
3_{.}
(l)
It is noted that the error onIz′z′ estimated by Eq. (l) increases rapidly with increasing values oft/a.

Thus, fort/a= 0.1, the error is about13%. The errors related to the estimated values ofIy′y′andIy′z′

*are somewhat smaller, i.e. about*5%and7%, respectively.

From now on, the origin of the(x, y, z)-coordinate system is placed at the bending centre. Then, the constitutive matrix given by Eq. (1–47) takes the form

C=E A 0 0 0 Iyy −Iyz 0 −Iyz Izz . (1–54)

As a result of this, an axial forceN no longer induces deformations in they- andz-directions, and the bending momentsMyandMzdo not induce axial displacements. However, the bending

momentMy will still induce displacements in they-direction in addition to the expected

dis-placements in thez-direction. Similarly, the bending momentMzinduces displacements in both

they- andz-directions.

**1.4.2**

**Determination of the principal axes**

In order to uncouple the bending deformations, so thatMywill only induce deformations in the z-direction, andMzonly deformations in they-direction, a new(x′, y′, z′)-coordinate system is

introduced with origin inBand rotated the angleϕaround thex-axis as shown in Fig. 1–13.

κy, My κz, Mz κ′ y, My′ κ′ z, Mz′ ϕ N=N′ y z y′ z′ B

**Figure 1–13**Rotation of coordinate system.

Let{N, My, Mz}and{N′, My′, Mz′} denote the components of the generalised stresses in

the(x, y, z)-coordinate system and the(x′_{, y}′_{, z}′_{)}_{-coordinate system, respectively. The two sets}
of generalised stresses are related as

σ_{=}Bσ′_{,} _{(1–55a)}
where
σ′_{=}
N′
M′
y
M′
z
, σ=
N
My
Mz
, B=
1 0 0
0 cosϕ −sinϕ
0 sinϕ cosϕ
. (1–55b)

Likewise, the components of the generalised strains in the two coordinate systems are denoted
as{ε′_{, κ}′

y, κ′z}and{ε, κy, κz}, respectively. These are related as

where
ε′_{=}
ε′
κ′
y
κ′
z
, ε=
ε
κy
κz
, B=
1 0 0
0 cosϕ −sinϕ
0 sinϕ cosϕ
. (1–56b)

The constitutive relation in the(x′_{, y}′_{, z}′_{)}_{-coordinate system reads}

σ′_{=}_{C}′_{ε}′_{,} _{C}′_{=}_{E}
A 0 0
0 Iy′y′ −Iy′z′
0 −Iy′z′ Iz′z′
. (1–57)

The corresponding constitutive relation in the(x, y, z)-coordinate system is given by Eq. (1–46) withCgiven by Eq. (1–54). Use of Eqs. (1–55a) and (1–56a) in Eq. (1–46) provides

Bσ′_{=}_{C}_{ε}_{=}_{CB}_{ε}′ _{⇒} _{σ}′_{=}_{B}T_{CB}_{ε}′_{,} _{(1–58)}

where it has been utilised thatB−1_{=}_{B}T_{. Comparison of Eq. (1–57) and Eq. (1–58) leads the}

following relation between the constitutive matrices
C′ _{=}_{B}T_{CB}
=E
1 0 0
0 cosϕ sinϕ
0 −sinϕ cosϕ
A 0 0
0 Iyy −Iyz
0 −Iyz Izz
1 0 0
0 cosϕ −sinϕ
0 sinϕ cosϕ
=E
A 0 0
0 C′
22 C23′
0 C′
32 C33′
. (1–59a)
where
C′

22= cos2ϕIyy−2 cosϕsinϕIyz+ sin2ϕIzz, (1–59b)

C′

23=C32′ =−sinϕcosϕ(Iyy−Izz)−(cos2ϕ−sin2ϕ)Iyz, (1–59c) C′

33= cos2ϕIzz+ 2 cosϕsinϕIyz+ sin2ϕIzz. (1–59d)

From Eq. (1–57) and Eq. (1–59) follows that Iy′y′=

1

2(Iyy+Izz) + 1

2(Iyy−Izz) cos(2ϕ)−Iyzsin(2ϕ), (1–60a)
Iy′_{z}′ =−
1
2sin(2ϕ)(Iyy−Izz)−cos(2ϕ)Iyz, (1–60b)
Iz′z′ =
1
2(Iyy+Izz)−
1
2(Iyy−Izz) cos(2ϕ) +Iyzsin(2ϕ), (1–60c)

where use has been made of the relations

sin(2ϕ) = 2 sinϕcosϕ, cos(2ϕ) = cos2_{ϕ}_{−}_{sin}2_{ϕ,}

cos2ϕ= 1

2(1 + cos 2ϕ), sin

2_{ϕ}_{=} 1

2(1−cos

Uncoupling of bending deformations in the(x′_{, y}′_{, z}′_{)}_{-coordinate system requires that}_{I}

y′_{z}′ = 0.

This provides the following relation for the determination of the rotation angleϕ:

−1 2sin(2ϕ)(Iyy−Izz)−cos(2ϕ)Iyz= 0 ⇒ tan(2ϕ) = 2Iyz Izz−Iyy for Iyy6=Izz, cos(2ϕ) = 0 for Iyy=Izz. (1–61)

Note thatcos(2ϕ) = 0implies thatsin(2ϕ) =±1. The sign ofsin(2ϕ)is chosen as follows:

sin(2ϕ) = 1 ⇒ ϕ= 1_{4}π for Iyz <0
sin(2ϕ) =−1 ⇒ ϕ= 3_{4}π for Iyz >0
. (1–62)

Then, Eq. (1–60) provides the following solutions forIy′y′andIz′z′:

Iy′y′=

1

2(Iyy+Izz)+|Iyz |, Iz′z′ = 1

2(Iyy+Izz)− |Iyz|. (1–63)

ForIyy6=Izzthe solution fortan(2ϕ)is fulfilled for the following two alternative solutions

forsin(2ϕ)andcos(2ϕ):

sin(2ϕ) =−2Iyz J , cos(2ϕ) = Iyy−Izz J , (1–64a) sin(2ϕ) = 2Iyz J , cos(2ϕ) =− Iyy−Izz J , (1–64b) where J =q(Iyy−Izz)2+ 4Iyz2 . (1–65)

The sign definition in Eq. (1–64a) is chosen. This implies that

2ϕ∈[0, π] for Iyz<0 and 2ϕ∈[π,2π] for Iyz>0. (1–66)

Insertion of the solution for sin(2ϕ)andcos(2ϕ) into Eq. (1–60) provides the following results forIy′y′ andIz′z′:

Iy′y′= 1 2(Iyy+Izz) + 1 2 (Iyy−Izz)2+ 4Iyz2 J , (1–67a) Iz′z′ = 1 2(Iyy+Izz)− 1 2 (Iyy−Izz)2+ 4Iyz2 J , (1–67b)

or, by insertion ofJ, cf. Eq. (1–65), Iy′y′= 1 2(Iyy+Izz) + 1 2 q (Iyy−Izz)2+ 4Iyz2 , (1–68a) Iz′z′ = 1 2(Iyy+Izz)− 1 2 q (Iyy−Izz)2+ 4Iyz2 . (1–68b)

(a) (b) ϕ ϕ y y z z y′ y′ z′ z′ x=x′ x=x′ B B

**Figure 1–14**Position of principal axes: (a)Iyz<0and (b)Iyz>0.

The coordinate axesy′_{and}_{z}′_{are known as the principal axes of the cross-section, whereas}_{I}_{y}_{′}_{y}_{′}
andIz′z′*are called the principal moments of inertia. It follows from Eqs. (1–63) and (1–67) that*

the choices of signs forsin(2ϕ)implies thatIy′y′ becomes the larger of the principal moments

of inertia andIz′z′ is the smaller principal moment of inertia. It is emphasised that this choice

is performed merely to have a unique determination ofϕ. Three other choices ofϕare possible
obtained by additional rotations of the magnitudesπ_{2},πand 3_{2}πrelative to the indicated.

If the cross-section has a symmetry line, and they-axis is placed along this line, thenIyz = 0.

Hence, a symmetry line is always a principal axis. Since the principal axes are orthogonal, the z-axis is also a principal axis—even if the cross-section is not symmetric around the axis.

**Example 1.2 Determination of principal axes coordinate system**

The cross-section analysed in Example 1.1 is reconsidered. The thin-wall approximation is used, so the moments of inertia are given by Eq. (l) in Example 1.1 and repeated here (without the primes):

Iyy≃ 233 84ta 3 ≈2.7738·ta3, Izz≃ 92 21ta 3 ≈4.3810·ta3, Iyz ≃ − 11 7ta 3 ≈ −1.5714·ta3.(a) The position of the bending centre relatively to the top-left corner of the cross-section is provided in Fig. A. From Eq. (a) and Eq. (1–65) follows that

J=
s
233
84ta
3_{−}92
21ta
3
2
+ 4
−11
7 ta
3
2
=ta3·
√
9769
28 , (b)

which by insertion into Eq. (1–64) provides:

sin(2ϕ) =−2· −
11
7ta
3
·28
ta3_{·}√_{9769} =
88
√
9769 ≈0.8903, (c)
cos(2ϕ) =
233
84ta
3_{−}92
21ta
3
·28
ta3_{·}√_{9769} =−
3780
84·√9769 ≈ −0.4553. (d)
*(continued)*

y
z
ϕ
y′
z′
x=x′
t
t
2t
a
2a
2a
B
yB≃ 4_{7}a
zB≃_{14}9a

**Figure A** Position of principal axes coordinated system for the thin-walled cross-section.

From Eqs. (c) and (d) it is found thatϕ = 1.0217 radians corresponding toϕ = 58.5418◦_{. Hence,}

ϕ∈[0,π

2]in agreement withIyz=− 11

7ta
3_{<}_{0}_{.}

*Finally, the moments of inertia in the principal axes coordinate system follow from Eq. (1–67), i.e.*
Iy′y′
Iz′z′
=
1
2
233
84 +
92
21
±1_{2}
s
233
84 −
92
21
2
+ 4
11
7
2
ta
3_{=}
5.3423ta3_{,}
1.8124ta3_{.}
(e)

Clearly,Iy′y′ is greater than any ofIyyorIzz, whereasIy′y′ is smaller than the bending moments of

inertia defined with respect to the originaly- andz-axes.

**1.4.3**

**Equations of equilibrium in principal axes coordinates**

From now on it will be assumed that the(x, y, z)-coordinate system forms a principal axes coor-dinate system with origin at the bending centre. In this case, the system of differential equations (1–28) for a Timoshenko beam uncouples into three differential subsystems. Thus, the axial deformation is governed by the equation

d dx EAdwx dx +qx= 0, (1–69)

whereas bending deformation in they-direction is defined by the coupled equations d dx EIzdθz dx +GAz dwy dx −θz +mz= 0, (1–70a) d dx GAy dwy dx −θz +qy = 0, (1–70b)

where the double indexyyon the bending moment of inertia has been replaces by a single index yin order to indicate that the principal-axes coordinates are utilised.

Similarly, the flexural deformations in thez-directions are determined by
d
dx
EIydθy
dx
−GAz
dwz
dx +θy
+my= 0, (1–71a)
d
dx
GAz
_{dw}
z
dx +θy
+qz= 0, (1–71b)

where again the double index on the bending moment of inertia has been replaced by a single index. As seen from Eqs. (1–70) and (1–71), {wy, θz} and {wz, θy} are still determined by

pairwise coupled ordinary differential equations of the second order.

For a Bernoulli-Euler beam, the system of ordinary differential equations (1–33) uncouples completely into the following differential equations for the determination ofwx,wyandwz:

d
dx
EAdwx
dx
+qx= 0, (1–72a)
d2
dx2
EIz
d2_{w}
y
dx2
−qy+
dmz
dx = 0, (1–72b)
d2
dx2
EIy
d2_{w}
z
dx2
−qz−
dmy
dx = 0. (1–72c)

**Example 1.3 Plane, fixed Timoshenko beam with constant load per unit length**

Figure A shows a plane Timoshenko beam of the lengthlwith constant bending stiffnessEIzand shear stiffnessGAy. The beam is fixed at both end-sections and is loaded with a constant loadqyand a constant moment loadmz. The displacementwy(x), the rotationθz(x), the shear forceQy(x)and the bending moment are to be determined.

x y z l mz δqy EIzz, GAy

**Figure A** Fixed beam with constant load per unit length.

The differential equations for determination ofwy(x)andθz(x)follow from Eq. (1–70). Thus,

EIz
d2_{θ}
z
dx2 +GAy
_{dw}
y
dx −θz
+mz= 0,
d
dx
GAy
_{dw}
y
dx −θz
+qy= 0. (a)
According to Eq. (1–35), the boundary conditions are:

wy(0) =wy(l) = 0, θz(0) =θz(l) = 0. (b)

Integration of the second equation in Eq. (a) provides: Qy=GAy dwy dx −θz =−qyx+c1 (c)

Then, the following solution is obtained forθz(x)from the first equation in Eq. (a):

EIzd
2_{θ}
z
dx2 =qyx−(c1+mz) ⇒ EIzθz(x) =
1
6qyx
3
−1
2(c1+mz)x
2_{+}_{c}
2x+c3. (d)
Further, the boundary conditionsθz(0) =θz(l) = 0provide

c3= 0, c2=−

1 6qyl

2_{+}1

2(c1+mz)l. (e)

Hence, the following reduced form is obtained forθz(x):

θz(x) =

1 6EIz

qy(x3−xl2)−3(c1+mz)(x2−xl). (f) Next, Eq. (f) is inserted into Eq. (c) which is subsequently integrated with respect tox, leading to the following solution forwy(x):

GAydwy
dx =−qyx+c1+
GAy
6EIz qy(x
3
−xl2)−3(c1+mz)x2−xl) ⇒
GAywy(x) =−1
2qyx
2
+c1x+c4+ GAy
6EIz
1
4qy(x
4
−2x2l2)−1_{2}(c1+mz)(2x3−3x2l)
. (g)
The boundary conditionswy(0) =wy(l) = 0provide the integration constants

c4= 0, c1= 1 2qyl− 1 Φy+ 1 mz, (h) where Φy= 12 EIz GAyl2 (i) Then, Eq. (a) and Eq. (f) provide the following solutions:

wy(x) = qy
2GAy
(l−x)x+ qy
24EIz
(l−x)2x2−_{GA}mz
y
x
Φy+ 1−
mz
12EIz
Φy
Φy+ 1
(2x3−3x2l) ⇒
wy(x) =
qy
24EIz
(l−x)2x2+ Φyl2(l−x)x
−_{12}m_{EI}z
z
Φy
Φy+ 1
(2x3−3x2l+xl2), (j)
θz(x) = qy
12EIz
(2x3−3x2l+xl2) + mz
2EIz
Φy
Φy+ 1
(l−x)x. (k)

The non-dimensional parameterΦy is a measure of the influence of the shear deformations. For a
rectangular cross-section with the heighthwe haveIz= _{12}1h2AandAy=5_{6}A. ThenΦybecomes

Φy= 72 5 · h2 l2 E G. (l)

Hence, shear deformations are primarily of importance for short and high beams. On the other hand, for
*long beams with a small height of the cross-section, shear deformations are of little importance, i.e. only*
the bending deformation is significant. *(continued)*

For Bernoulli-Euler beams we haveγxy= 0, corresponding toGAy=∞, cf. Eqs. (1–16) and (1–27). Then,Φy= 0and Eqs. (j) and (k) reduce to

wy(x) = qy 24EIz (l−x)2x2, (m) θz(x) = qy 12EIz (2x3−3x3l+xl). (n)

It is remarkable that the distributed moment loadmzdoes not induce any displacements or rotations in the considered beam with Bernoulli-Euler kinematics.

The shear forceQy(x)and bending momentMz(x)follow from Eq. (c), Eq. (h) and Eq. (k),
*respec-tively, i.e.*
Qy(x) =−qyx+c1=1
2qy(l−2x)−
1
Φy+ 1mz, (o)
Mz(x) =EIz
dθz
dx =
qy
12(6x
2
−6xl+l2) +mz
2
Φy
Φy+ 1
(l−2x). (p)

For a Bernoulli-Euler beam these results reduce to Qy(x) = 1 2qy(l−2x)−mz, (q) Mz(x) = qy 12(6x 2 −6xl+l2). (r)

The constant moment loadmzonly induces a constant shear force of magnitude−mz, whereasMz(x) is not affected by this load. Especially, forx= 0andx=l, Eq. (o) and Eq. (p) provide

Qy(0) =
1
2qyl−mz, My(0) =
1
12qyl+
1
2
Φy
Φy+ 1
mzl, (s)
Qy(l) =−1
2qyl−mz, My(l) =
1
12qyl
2
−1_{2}_{Φ}Φy
y+ 1
mzl. (t)

The displacement at the midpointx=l/2follows from Eq. (k): wy(l/2) =

qyl4

384EIz

(1 + 4Φy). (u)

The first and second terms within the parenthesis specify the contributions from bending and shear, re-spectively. Again the parameterΦyreveals itself as a measure of the relative contribution from shear

deformations.

**1.5**

**Normal stresses in beams**

For at beam without warping, the normal stressσxx(x, y, z)in terms of the generalised strains follows from Eqs. (1–17), (1–20) and (1–24):

σxx=E(ε−κzy+κyz). (1–73)

In the principal axes coordinate system, whereSy = Sz = Iyz = 0, the generalised strains {ε, κy, κz}are related to the conjugated generalised stresses{N, My, Mz}as determined from

Eqs. (1–45a), (1–45b), (1–46) and (1–47) forSy =Sz=Iyz = 0*, i.e.*
ε= N
EA, κy=
My
EIy
, κz=
Mz
EIz
. (1–74)

Insertion of Eq. (1–74) into Eqs. (1–19) and (1–24) provides the result for the axial stress in terms of the generalised stresses,

σxx= N
A −
Mz
Iz y+
My
Iy z.
(1–75)
*Equation (1–75) is due to Navier, and is therefore referred to as Navier’s formula. It should be*
noticed that Eq. (1–75) presumes that the stresses are formulated in a principal axes coordinate
system, soIy andIz indicate the principal moments of inertia. The relation is valid for both

Timoshenko and Bernoulli-Euler beams. This is so because only the relation (1–15), but not the relation (1–18) has been utilised. Hence, Eq. (1–75) is based on the assumption that plane cross-sections remain plane, but not that they remain orthogonal to the beam axis.

*The so-called zero line specifies the line in the*(y, z)-plane on whichσx= 0. The analytical

expression for the zero line becomes N A − Mz Iz y+ My Iy z= 0. (1–76) It is finally noted that warping introduces displacements in the axial direction in addition to those provided by bending. However, if the torsion is homogeneous, these displacements will not introduce any normal strains and therefore no normal stresses. Hence, Navier’s formula is also valid in the case of St. Venant torsion, but in the case of Vlasov torsion, or inhomogeneous torsion, additional terms must be included in Eq. (1–75).

**1.6**

**The principle of virtual forces**

*In this section the principle of virtual forces is derived for a plane Timoshenko beam of the length*
l. The deformation of the beam is taking place in the(x, y)-plane. In the referential state, the
left end-section is placed at the origin of the coordinate system and thex-axis is placed along the
bending centres of the cross-sections, see Fig. 1–15.

*The principle of virtual forces is the dual to the principle of virtual displacements. In the*
principle of virtual displacements the actual sectional forces and sectional moments are assumed
to be in equilibrium with the loads and the reaction forces applied at the end sections. The virtual
displacements and rotations are considered as arbitrary increments to the actual displacements
and they only need to fulfil homogeneous kinematic boundary conditions, so that the combined
field made up by the actual and the virtual fields always fulfils the actual non-homogeneous
boundary conditions as given by Eq. (1–35). Further, the generalised virtual strains defining the
internal virtual work must be derived from the virtual displacement and rotation fields.

In contrast, the principle of virtual forces presumes that the displacements and rotations of the beam are fulfilling the kinematic boundary conditions, and that the generalised internal strains are compatible to these fields. The actual loads on the beam are superimposed with the virtual incremental loads per unit lengthδqx andδqy, the virtual moment load per unit lengthδmz,

x y z l δqx δqy δmz δMz,1 δMz,2 δQy,1 δN1 δN2 δMz δQy δN δQy,2

**Figure 1–15**Virtual internal and external forces.

the virtual reaction forcesδNj andδQy,j along thex- andy-directions, and the virtual reaction

momentsδMzin thez-directions, wherej = 1andj= 2indicate the left and right end-sections,

respectively. Due to the load increments, the internal section forces and section moment achieve incrementsδN, δQy and δMz, see Fig. 1–15. These variational fields are assumed to be in

equilibrium with the variational load fieldsδqx, δqyandδmz, and to comply with the variations δNj, δQy,j andδMz,j of the reaction forces and reaction moments. In what follows,δN,δQy

andδMz will be referred to as the virtual internal forces, whereasδqx,δqy,δmz,δNj,δQy,j

andδMz,jare called the virtual external forces.

The starting point is taken in the kinematical conditions provided by Eqs. (1–12) and (1–17) and rewritten in the form

dwx dx −ε= 0, dwy dx −θz−γxz= 0, dθz dx −κz= 0. (1–77)

The virtual internal forces are related to the virtual external loads per unit lengthδqx,δqy and δmzvia the following equations of equilibrium, cf. Eqs. (1–4a) and (1–4b):

d(δN) dx +δqx= 0, d(δQy) dx +δqy= 0, d(δMz) dx +δQy+δmz = 0. (1–78) The first equation in Eq. (1–78) is multiplied withδN(x), the second equation is multiplied withδQy(x), and the third equation withδM(z). Next, the equations are integrated fromx= 0

tox=l, and the three resulting equations are added, leading to the identity
Z l
0
δN
_{dw}
x
dx −ε
+δQy
_{dw}
y
dx −θz−γxy
+δMz
_{dθ}
z
dx −κz
dx= 0 (1–79)

Integration by parts is carried out on the first terms within the innermost parentheses, leading to
δN wx+δQywy+δMzQz
l
0
−
Z l
0
_{d}_{(}_{δN}_{)}
dx wx+
d(δQy)
dx wy+
d(δMz)
dx θz−δQyθz
dx
=
Z l
0
δN ε+δQy·γxy+δMz·κz
dx. (1–80)

Upon utilisation of Eq. (1–78), this is reduced to h δN·wx+δQy·wy+δMz·θz il 0+ Z l 0 δqx·wx+δqy·wy+δmz·θz dx = Z l 0 δN·ε+δQy·γxy+δMz·κz dx. (1–81)

The generalised strains on the right-hand side of Eq. (1–81) are now expressed in mechanical quantities by means of Eq. (1–74). Further,δN,δQy andδMz fulfil the following boundary

conditions atx= 0andx=l, cf. Fig. 1–15,

δN(0) =−δN1, δQy(0) =−δQy,1, δMz(0) =−δMz,1, (1–82a)

δN(l) = δN2, δQy(l) = δQy,2, δMz(l) = δMz,2. (1–82b) Equation (1–81) then obtains the following final form:

2 X j=1 δNj·wx,j +δQy,j·wy,j+δMz,j·θz,j + Z l 0 δqx·wx+δqy·wy+δmz·θz dx = Z l 0 δN·N EA + δQy·Qy GAy +δMz·Mz EIz dx, (1–83)

wherewx,j,wy,jandθz,jdenote the displacements in thex- andy-directions and the rotation

in thez-direction at the end-sections, respectively. Equation (1–83) represents the principle of virtual forces. The left- and right-hand sides represent the external and internal virtual work, respectively.

The use of Eq. (1–83) in determining the displacements and rotations of a Timoshenko beam is demonstrated in Examples 1.4 and 1.5 below. Furthermore, the principle of virtual forces may be used to derive a stiffness matrix for a Timoshenko beam element as shown later.

**Example 1.4 End-displacement of cantilevered beam loaded with a force at the free end**

Figure A shows a plane Timoshenko beam of the lengthlwith constant axial stiffnessEA, shear stiffness GAyand bending stiffnessEIz. The beam is fixed at the left end-section and free at the right end-section, where it is loaded with a concentrated forceQy,2in they-direction. The displacementwy,2at the free end is searched.

The principle of virtual forces Eq. (1–83) is applied with the following external virtual loads:δqx =

δqy =δmz = 0,δN1 =δQy,1 =δMz,1=δN2 =δMz,2 = 0andδQy,2 = 1. FurtherN(x) = 0. Then, Eq. (1–83) reduces to

1·wy,2=
Z l
0
δQy·Qy
GAy +
δMz·Mz
EIz
dx. (a)
*(continued)*

1 x x y y z z l l l Mz Qy δMz δQy Qy,2·ℓ Qy,2 Qy,2 δQy,2= 1 EA, GAy, EIz

**Figure A** Fixed plane Timoshenko beam loaded with a concentrated force at the free end: Actual force and section
forces (left) and virtual force and section forces (right).

The variation of the bending momentMz(x)and the shear forceQy(x)from the actual loadQy,2has been shown in Fig. A on the left. The corresponding variational moment fieldδMz(x)and shear force

δQy(x)fromδQy,2 = 1are shown in Fig. A on the right. Insertion of these distributions in Eq. (b) provides the solution

wy,2=Qy,2l GAy +1 3 Qy,2l3 EIz = 1 12(4 + Φy)· Qy,2l3 EIz , (b)

whereΦyis given by Eq. (i) in Example 1.3. The deformation contributions from shear and bending are additive. This is a consequence of the additive nature of the flexibilities indicated by Eq. (1–83) in contribution to the fact that the beam is statically determinate, which provides the fieldsMz(x)and

Qy(x)as well asδMz(x)andδQy(x)directly.

**Example 1.5 End-deformations of fixed beam loaded with a moment at the free end**

The beam described in Example 1.4 is considered again. However, now the free end is loaded with a concentrated momentMz,2. The displacementwy,2 and the rotationθz,2 of the end-section is to be found.

At the determination ofwy,2fromMz,2, the principle of virtual forces given by Eq. (1–83) is again applied withδQy,2 = 1and all other external variational loads equal to zero, leading to Eq. (a) in Example 1.4. However,Mz(x)andQy(x)are now caused byMz,2, and are given as shown in Fig. A, whereasδMz(x)andδQy(x)are as shown in Fig. A of Example 1.4. Then,wy,2becomes

wy,2=
Z l
0
δMz·Mz
EIz
dx=1
2
My,2l2
EIz
. (a)
*(continued)*

1 x x y y z z l l Mz δMz EA, GAy, EIz Mz,2 Mz,2 Qy= 0 δQy= 0 δMz,2= 1

**Figure A** Fixed plane Timoshenko beam loaded with a moment at the free end: Actual moment and section forces
(left) and virtual moment and section forces (right).

At the determination ofθz,2 the principle of virtual forces Eq. (1–83) is applied with the following external virtual loadsδqx =δqy = δmz = 0, δN1 =δQy,1 = δMz,1 =δN2 = δQy,2 = 0and δMz,2= 1. Then, Eq. (1–83) reduces to

1·θz,2= Z l 0 δQy·Qy GAy +δMz·Mz EIz dx. (b)

The variation ofQy(x)andMz(x)fromMy,2has been shown in Fig. A on the left, and the variation of δQy(x)andδMz(x)fromδMz,2= 1is shown in Fig. A on the right. Thenθz,2becomes

θz,2= Z l 0 1·Mz,2 EIz dx= Mz,2l EIz . (c)

In the present load case, the shear force is given asQy(x) = 0. Consequently it will not induce any

contributions in Eq. (a) and Eq. (c).

**1.7**

**Elastic beam elements**

When frame structures consisting of multiple beams are to be analysed, the establishment of
analytical solutions is not straightforward and instead a numerical solution must be carried out.
*For this purpose, a discretization of the frame structure into a number of so-called beam *

*ele-ments is necessary, eventually leading to a finite-element model. The aim of the present section*

is not to provide a full introduction to the finite-element method for the analysis of frame
*struc-tures, e.g. tower blocks with a steel frame as the load-carrying structure. However, a *
formula-tion is given for a single beam element to be applied in such analyses. Both the Timoshenko
and Bernoulli-Euler beam theories are discussed in this context, and plane as well as
three-dimensional beams are touched upon.

**1.7.1**

**A plane Timoshenko beam element**

Firstly, the stiffness matrix and element load vector is derived for a plane Timoshenko beam ele-ment with constant axial stiffnessEA, shear stiffnessGAyand bending stiffnessEIz, cf. Fig. 1–

16. The stiffness relation is described in an(x, y)-coordinate system with origin at the left end-section and thex-axis along the bending centres.

x y z l EA, GAy, EIz Qy,1, wy,1 N1, wx,1 Mz,1, θz,1 Qy,2, wy,2 N2, wx,2 Mz,2, θz,2 1 2

**Figure 1–16**Plane Timoshenko beam element with definition of degrees of freedom and nodal reaction forces.
At the end-nodes, nodal reaction forcesNjandQy,jare acting along thex- andy-directions,

respectively, and reaction momentsMz,jare applied around thez-axis. Here,j = 1andj = 2

stand for the left-end and right-end nodes of the beam element, respectively, and the reaction forces and moments are in equilibrium with the remaining external loads on the element for arbitrary deformations of the beam.

The element has 6 degrees of freedom defining the displacements and rotations of the end-sections, cf. Fig. 1–16. These are organised in the column vector

we =
we_{1}
we_{2}
=
wx,1 wy,1 θz,1 wx,2 wy,2 θz,2
T
(1–84)
The sub-vectorwejdefines the degrees of freedom related to element nodej.

Similarly, the reaction forcesNj,Qy,jandMz,j,j = 1,2, at the end-sections, work

conju-gated towx,j,wy,jandθz,j, are stored in the column vector

re=
re_{1}
re2
=
N1 Qy,1 Mz,1 N2 Qy,2 Mz,2
T
(1–85)
x
y
z
mz
px
py
Qy,1
N1
Mz,1
Qy,2
N2
Mz,2

**Figure 1–17**External loads and reaction forces from external loads on a plane beam element.

The equilibrium of the beam element relating the nodal reaction forces to the degrees of freedom of the element may be derived by the principle of virtual displacements as demonstrated

in a subsequent paper. The resulting equilibrium equations on matrix form may be written on the form

re=Kewe+fe. (1–86)

The vectorfein Eq. (1–86) represents the nodal reaction forces from the external element loads

whenwe=0, i.e. when the beam is fixed at both ends as shown in Fig. 1–17. We shall merely

consider constant element loadsqxandqyper unit length in thex- andy-directions, and a

con-stant moment load per unit lengthmzin thez-direction, see Fig. 1–17. The reaction forces and

reaction moments follow from Eqs. (s) and (t) in Example 1.3:

fe=
−1_{2}qxl
−1_{2}qyl+mz
−_{12}1qyl2−_{2}1_{Φ}Φ_{y}_{+1}y mzl
−1_{2}qxl
−1_{2}qyl−mz
1
12qyl2−
1
2
Φy
Φy+1mzl
. (1–87)

The matrixKein Eq. (1–86) denotes the stiffness matrix in the local(x, y, z)-coordinate system.

Letwidenote theith component ofwe. Then, theith column inKerepresents the nodal reaction

forces forfe =0, and withwi = 1andwj = 0, j 6= i. These forces are obtained following

the derivations in Example 1.3 from Eq. (a) to Eq. (t) withqy=mz= 0and with the boundary

condition in Eq. (b) replaced by the indicated conditions. Because of the symmetry of the prob-lem, only two such analyses need to be performed. Still, this is a rather tedious approach. Partly because of this, and partly in order to demonstrate an alternative approach, the stiffness matrix will be derived based on the principle of virtual forces.

x x y y wx wy θz Undeformed state Undeformed state (a) (b)

**Figure 1–18**Rigid-body modes of a plane beam element: (a) Translation and (b) rotation.

The beam element has 6 degrees of freedom, by which a total of 6 linear independent modes of deformation may be defined. These consist of 3 linear independent rigid body modes and 3 linear independent elastic modes. The rigid modes may be chosen as a translation in the x-direction, a translation in the y-direction and a rotation around the z-direction as shown in Fig. 1–18. Any rigid body motion of the beam element may be obtained as a linear combination of these component modes of deformation. Obviously, the rigid body motions do not introduce stresses in the beam. Hence, the axial forceN, the shear forceQyand the bending momentMz

are all zero during such motions.

Since, axial elongations are uncoupled from bending deformations, the elastic elongation mode is uniquely defined as shown in Fig. 1–19a. The two bending deformation modes may be